Another way to do this is to start with a regular pentagon where the length of each side is one unit. I'll outline the steps. (You can fill in the details.)
Call the pentagon ABCDE. Draw diagonals AC and BE, which intersect at point F. Using the fact that each angle of the pentagon is 108 degrees, you can prove the triangles BFC and AFE are isosceles. Therefore the lengths of line segments CF and EF are also 1 unit. Also note the angles for the triangles are 72-72-36.
Now show that ABF and CAB are similar triangles. Since they are similar, this means BE/AB = AB/BF. Since BE = EF + BF, substitute to get (EF + BF)/AB = AB/BF.
Let's define x to be the length of BF. Then, since EF = 1, the equation becomes (1 + x)/1 = 1/x, or x^2 + x - 1 = 0. Use the quadratic equation to get x = (sqrt(5) - 1) / 2.
Finally, let's create an 18-72-90 by drawing the altitude of triangle BFC from vertex C to side BF. Since BC = FC, the altitude hits BF at the midpoint of BF, which we will call point G. So, (a) triangle GBE is an 18-72-90 triangle, (b) BG = BF/2 = x/2.
Using this 18-72-90 triangle, sin 18 degrees = BG / BC = x/2. Therefore sin 18 = (sqrt(5) - 1) / 4.